units in indefinite integral

Question

When answering this question, I came up with the following:
Suppose we have a length $l$ and integrate $\frac 1 l$ over $l$: $$\int \frac {\operatorname d\!l}l=\ln l$$ Since the dimension of $l$ is length, and $\ln$ should have a dimensionless argument, this cannot be a 'good' integral in the physical sense, although it would be mathematically (omitting units). A way to solve this would be to reformulate the indefinite integral like this: for $c=-\ln (1 \text{meter})$. $$ \int \frac {\operatorname d\!l}l=\ln l+C=\ln l-\ln( 1\text{m})=\ln\left(\frac{l}{1\text{meter}}\right) $$ The question is whether the first integral given would be acceptable and how to deal with the two logarithms with bad arguments, which, when combined, aren't a problem anymore.

Answer 1

I think your integral does not make any sense, at least in a physical sense, because it's not a definite integral. Indeed, when computing the non-dimensional elongation in a given direction, this should be as follows:

$$\epsilon = \int^L_{L_0} \frac{d l}{l} = \ln{\frac{L}{L_0}},$$

so the units are correct. This also happens when considering the work done by a force along a given path. For example, considering the dissipated energy due to friction along a straight line of length $x-x_0$ it gives:

$$W_r = \int^x_{x_0} F_r \, dx = F_r (x - x_0),$$

assuming $F_r$ is constant, which has units of work or energy, [J] (there's a lot of more interesting examples). In resume, in physics/engineering you almost never work with indefinite integrals in a physical meaningful way.

Cheers!

Answer 2

The indefinite integral is never equal to a particular function, so both suggestions you gave are incorrect. What we do have, however, is that the indefinite integral is a class of functions that differ by a constant. When you use the indefinite integral to compute a definite integral, the constant will vanish, and so will any units that got fed into the logarithm, if indeed the original integral made sense with the units. For instance, $\ln(2\text{ m}) - \ln(1\text{ m}) = ln(2)-ln(1)$.